where x and y are
orthogonal coordinates on a plane. The conic is a parabola if and only if B2
= 4AC, in which case we can define constants a,c such that A = a2
and C = c2 and write the equation in the form

Given four arbitrarily
selected points Pj on the plane, with coordinates xj,yj,
for j = 0,1,2,3, we can shift all the points so that P0 lies on the
origin. The parabola through those four
points then passes through the origin, so we have F = 0. Also, since we
can divide through by any of the quantities a2, c2, D,
E (if they are non-zero) there are really only three degrees of freedom,
which we can use to require that the remaining three points (Pj,
j=1,2,3) satisfy the equation, so we have

Solving any two of these
for D and E, and substituting into the third, we find the necessary and
sufficient condition for the parameters a,c to give a parabola through these
points is

Letting X and Y
denote vectors with components xi and yi respectively,
for any given set of three points, we can define the cross-product vector V
= XxY with the components

This establishes a mapping
between three-point configurations (equivalent up to translations and
rotations) and vectors in a three-dimensional space. These quantities are
invariant under rotations as well as translations. Also, if we divide through
the equation by c2 we get a quadratic in the ratio r = a/c. Letting uj = rxj + yj,
and letting Q denote the vector with components uj2, the
preceding equation can then be written as the vanishing of the dot product V∙Q
= 0. Hence the necessary and sufficient condition for a and c to give a
parabola through the four points is that Q must be in the same plane
as X and Y.

The quadratic in r can also be written as

The two roots are

We can also define the
one-parameter family of vectors q(s)with
components

By varying the value of s we vary the direction of this vector. The two roots r are such that q(r) is orthogonal to
V, i.e., such that q(r)×V = 0.
These two values correspond to the two orientations for which a normal
parabola through the origin passes through all three of the points. The
equation of this parabola is

where r is either of the two roots of the characteristic quadratic and the
values of the coefficients D and E are given by

As an example, consider the
four co-plane points with coordinates

The characteristic polynomial
is

with the roots

From this we find the
corresponding values of the D and E coefficients

Thus the two parabolas
through the four given points (including the origin) have the equations

where the ± are either all plus or all minus. Expanding the square and
collecting terms, we can split the equation into rationally incomensurate
parts (for rational x and y) as follows

This implies that any
rational points (x,y) on the parabola must also be on each of the curves
represented by the two expressions in parentheses. By Bezout's Theorem these
two curves intersect in four points, so the original four points (0,0), (-4,4),
(-3,2) and (3,1) are the only rational points on the parabola. Incidentally,
the curves given by the two parenthetical expressions are an ellipse and a
hyperbola respectively. These are shown in the figure below.

We can regard the
quantities axj + cyj as comprising the values of a
transformed variable, and since only the ratio of a/c is important we can
parameterize them as a = sin(q) and c = cos(q). (It follows that the parameter r equals tan(q).) Then we have the transformed variable

In terms of this variable
the original equation of a parabola is

If we then define a second
variable consistent with the first by means of a simple plane rotation
through the angle q, we have

Thus the original variables
are given by

so we can substitute into
the parabola's equation to give

The preceding discussion
began with four arbitrary points on the plane and then we translated the
points so as to place one of them at the origin. This simplifies the
expressions, but tends to obscure the full symmetry of the solution. Any
four points can be fit them to the general parabola equation

Letting uj
denote rxj + yj we can substitute the
four sets of coordinates into this equation, solve any three of them for D,E,F,
and then substitute these expressions into the fourth equation to give a
homogeneous equation in the uj quantities. Thus we have

Expanding the determinants
and collecting terms, this gives

where vij = xiyj- xjyi. Notice that the
quantities

represent half the signed areas of the triangles with vertices Pi,
Pj, and Pk. In terms of these areas the previous
equation can be written in the form

Expanding the uj
expressions, we get a quadratic in the parameter r

The discriminant of this
equation, given by the square of the middle parenthetical quantity minus the
product of the first and third, is

which shows (as one would
expect) that the discriminant is invariant under rotations and translations.
In fact, expressing the v and A functions in terms of the x and y coordinates
and simplifying, we find that this expression factors as

Thus the two values of r that give parabolas through the four points are

This expression is
invariant under transposition of any two indices, since the effect of any
transposition is simply to negate the numerator and denominator.

Since the directions of the
axes of symmetry of the two parabolas through any given four points in a
plane) can be determined by solving a quadratic equation, it follows that
lines with these directions can be constructed geometrically by Euclidean
methods, i.e., using only straight-edge and compass. Of course, not all
four-point configurations correspond to parabolas whose algebraic expressions
have real and finite coefficients. Obviously if any one of the four points
is inside the triangle formed by the other three, then the two implied
parabolas and complex and identical, and the locus intersects with the real
plane only in the original four given points (assumed real). In this case
the real and imaginary parts of the parabolic equation can be taken as the
real coefficients of two real hyperbolas that intersect in the four points.
If any two of the lines connecting distinct pairs the four points are
parallel, then the implied parabola has its vertex at infinity, and is
represented graphically by those two parallel lines.

The condition for the
configuration of points to give real parabolas can be inferred from the
discriminant. The solution is purely real if and only if the argument of the
square root is positive. The argument is A243A134A142A123,
which is (one sixteenth of) the product of the signed areas of the four
triangles that can be formed from three of the four given points. The
quantity is easily seen to be invariant under permutations of the indices, so
we can number the four points arbitrarily. The signs of the individual areas
are positive (by convention) if and only if the indices proceed in the
counter-clockwise direction. All cases are one of the two types shown below.

For the convex arrangement on
the left, the loops 243, and 142 proceed in the counter-clockwise direction,
and the loops 134 and 123 proceed in the clockwise direction, so the product
of the signed areas is positive. On the other hand, for the non-convex
arrangement on the right, the loops 243, 134, and 142 are counter-clockwise,
and the loop 123 is clockwise, so the product of the signed areas is
negative.

For the convex case, i.e.,
when none of the four points A,B,C,D is inside the triangle formed by the
other three, we can construct (by Euclidean methods) lines parallel to the
axes of symmetry of the two parabolas as follows:

-Let e and f denote the
feet of the perpendiculars to CD from A and B respectively.

-Let g denote the
intersection of the line through D perpendicular to CD, and the line through
A perpendicular to AC.

-Let h denote the
intersection of the line Dg and the line through B perpendicular to BC.

-Let i denote the
intersection of the line Ae and the line through h parallel to Ag.

-Let j denote the
intersection of the line Ae and the line through B parallel to Ae.

-Let k denote the
intersection of the line Ae and the line through B parallel to AB.

-Locate the point L on Ae
such that the segment kL has the same length and direction as the segment Aj.

-Draw the circle Q whose
diameter is the segment iL.

-Let m and n denote the
intersections of the line through k perpendicular to Ae with the circle Q.

-Let o denote the
intersection of the line Bf with the line through L perpendicular to Ae.

-The lines mo and no are
parallel to the axes of symmetry of the two parabolas through A, B, C, and D.

This construction for a
typical configuration of points A,B,C D is illustrated in the figure below.

Pappus of Alexandria
(c. 300 AD) gave a construction of the general conic through five given
points in Book VIII of The Collection. Isaac Newton treated the same problem
in the Principia (1687), and then more explicitly in Arithmetica Universalis
(1707). In the latter he based the solution on finding the direction of the
axis of symmetry, so it can be compared directly with the discussion above. Newton's
construction, given any four points A,B,C,D in the plane, is as follows:

- Let O denote the intersection
of the lines AB and CD.

- Locate the point E on the
line AB such that the segment OE is equal in length to OB and has
the same sense as AO. (If O is between A and B then E is at B.)

- Locate the point F on the
line CD such that the segment OF is equal in length to OC and has the same
sense as DO. (If O is between C and D then F is at C.)

- Let G denote the circle
on the diameter AE.

- Let H denote the circle
on the diameter DF.

- Let I and J denote the
intersections of the line through O perpendicular to AB and the circle G.

- Let K and L denote the
intersections of the line through O perpendicular to CD and the circle H.

- Let M and N denote the
points on AB that are a distance OI from O.

- Let P and Q denote the
points on CD that are a distance OK from O.

- The quadrangle MPNQ is a
parallelagram whose sides are parallel to the axes of the two parabolas that
pass through the points A, B, C, and D.

This construction is
illustrated in the figure below.

Incidentally, if we let f1(x,y)
= 0 and f2(x,y) = 0 denote the equations of the two parabolas that
pass through four given points, then obviously the quadratic curve given by

where m is any constant.
Thus we have a one-parameter family of conics, each of which passes through
the four given points. In effect, the two parabolas are the basis elements
for this family. Letting r1 and r2 denote the two values of r characterizing the two parabolas, we can write their equations as

The general member of the
family has the equation

The discriminant of this
quadratic equation is

Thus we can find the conic
through four points with any specified discriminant, since the discriminant
is proportional to m. To illustrate, consider again the example
discussed above, based on the four points (0,0), (-4,4), (-3,2), (3,1). The
figure below shows one of the basis parabolas along with the conics
corresponding to m = 100 and m = -100. The other
basis parabola is approached as m goes to infinity.